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Question

If the activity of 108Ag is 3 micro curie, the number of atoms present in it are ( λ= 0.005 sec1)
  1. 2.2 x 105
  2. 2.2 x 106
  3. 2.2 x 104
  4. 2.2 x 107

A
2.2 x 106
B
2.2 x 107
C
2.2 x 104
D
2.2 x 105
Solution
Verified by Toppr

λ=0.005sec1
1 Micro curie =3.7×104dps
Assuming all Ag atoms to be radioactive, and each showing
Let no of atoms present =1 dps no
Given:
(dNdt)activity=λN03.7×3×104dps=0.005×N0
N0=2.2×107atoms

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